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10 Aug 2005

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How does the code return the rays (ie sides with only one vertex and a direction). From what I can tell VVertexB is just set to NaN. But how does one generate the rays from this information?






Hi! For (partly) infinite edges I provided two properties do make drawing easier: FixedPoint and DirectionVector. FixedPoint is either the (one) Vertex that exists or the middle between the two data points. DirectionVector is orthogonal to the connecting line between the data points (this is a property of all voronoi edges).
The secret to getting the direction of an infinite edge is in the voronoi algorithm itself  it provides edges with 'left' and 'right' data points. This gives a direct way to calculating the direction from the vertex. It goes in the direction where LeftData is left and RightData is right. Simple
Cheers, Ben





Do you mean that the Voronoi edge ray must start at VVertexA and must have the direction of the line made by the points LeftData and RightData (i.e. from LeftData to RightData comes first). If yes, then your algorithm is wrong. Just test it on a rectangle and you will see the result. If I misunderstood you, please explain how to get the rays in more details.
VladovsoftSoftware products for fitness and health club management, storehouses, shops and barcode generation.





They start in VVertexA and go orthogonal to the line from LeftData to RightData, so that LeftData is on the left and RightData is on the right.





Ok, thanks. But what about the vertices. I made a simple polygon generator to test the algorithm and I saw that the location of the vertex for 3 points (i.e. a triangle) is sometimes out of the triangle. But it should be the center of the inscribed circle, shouldn't it ?
VladovsoftSoftware products for fitness and health club management, storehouses, shops and barcode generation.





No, Voronoi Vertex of a triangle is the center of it's outer circle, which is only inside the triangle for max(angle)<90 degree. (see wiki[^]) B.





Ok, thanks for the info. I was planning to use Voronoi diagrams for finding the Maximum Inscribed Circle in a simple polygon, but I guess I should read some more information about its properties to determine if it will work for me.
VladovsoftSoftware products for fitness and health club management, storehouses, shops and barcode generation.





According to this, if I take RightDataLeftData, I get the vector from left to right. Then swapping X and Y and negating one gets an orthogonal vector. Taking the nonunknown point and adding the orthogonal vector should produce the correct ray, with LeftData on the left and RightData on the right. However in some cases it does not, and the ray is cast in the wrong direction, as if LeftData and RightData are in the wrong order.
Do you have a correction for me, or updated version of this code?





Hi, sorry, I dont have any free time at the moment... can you verify whether the left/right data points are actually wrong? B.





I solved this the same way many other implementations do  don't try to draw the unbounded edges, just contain the points you want to draw within a much larger triangle or square.
With this, I don't have any issues.





Hello BenDi, I tried to fix the ray problem many people seem to have. I verified the left/right data points and they really seem to be wrong. If I draw a green/blue line from RightData to LeftData (where the left part of the line is green and the right part is blue) you can see in the picture below, that the right/left side is sometimes wrong. https://www.dropbox.com/s/1o4zl2izmkuv96d/voronoi_ray_problem.png[^]
Do you have any idea where I could fix that problem? How do you decide which one is the left and which one the right point?
It would be kind if you could help me with this problem, or at least give me a hint. Drawing the rays correctly is really necessary for me and I can't skip them.
Peh





Hi! As you can see from the image, the left data point (as seen from the center of the VoronoiEdge) is the one with the lower X coordinate. Can you make the ray problem more precise? Then maybe I can help more. Cheers, B.





Hi, thanks for your interest in my problem. I understand what you say and you are right for the image I posted. I made a more precise one now.
P are points M are mids between two points blue lines point to right data green lines point to left data V is the voronoi point (VVertexA)
Rays are drawn by hand so far. Starting at V.
https://www.dropbox.com/s/fqxkg9im5e5bnmk/voronoi_ray_problem2.png[^]
My Problem is how to get the correct direction of the ray vector.





Ah, I remember.
The right solution was posted a long time by ckaut:
On an unbounded edge, only one point, let's call is 'center', is set. Comput a vector
diff = (left_data+right_data)/2  center
and paint a line from center to center + N * diff (where N is large enough to go out of your viewing area.
For this it does not matter if left and right are correctly assigned.
BTW: I haven't looked at the code in a while, but it seems like I added
VoronoiEdge.FixedPoint (=center) and VoronoiEdge.DirectionVector (=diff, normalized)
for exactly this purpose Sorry for not documenting it better.





Yes this looks like a very easy solution and this was what I already tried. But it's only the solution for the half of this problem.
If the center (V) is within the triangle of the three Points (P) then this way is right (correct for this problem[^]).
But if the center (V) is outside the triangle then it fails too (still fails in this problem[^]). Ray13 should be the other direction here.





Uh, that's true. Appearantly I did not see this case during debugging. This will probably need a bit more smarts in the Algorithm itself to correctly assign left and right data points, but I might not have time for that in the near future.
Feel free to give it a shot: in FortuneVoronoi.cs, ProcessDataEvent, left/right data is assigned, but after that there are some special cases for how to insert the new tree nodes. My guess is that left/right data points should be assigned based on the same if/else distinguation. Give it a try!





This was a nasty trap ;(
I saw this example (Visualization of the 2D Voronoi Diagram and the Delaunay Triangulation[^]) of an implementation of your algorithm. I downloaded both (yours and the example) and picked the wrong one. What I didn't see is that this example is using an old and buggy version of your code.
With the code from this article here I finally got the correct rays by drawing a line using the DirectionVector and the FixedPoint. I better only post a class below how I did it. If anyone is interested in my full example code you can email me, because I don't want anyone to run into the same problems again by republishing another version of this code.
Thanks a lot for your help.
using System;
using System.Collections.Generic;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Windows.Forms;
using BenTools.Mathematics;
namespace FortuneVoronoiVisualisation
{
class Visualisation
{
public static Bitmap GetVoronoiMap(int width, int height, HashSet<Vector> datapoints)
{
Bitmap bmp = new Bitmap(width, height);
VoronoiGraph graph = Fortune.ComputeVoronoiGraph(datapoints);
Graphics g = Graphics.FromImage(bmp);
for (int w = 0; w <= width; w = w + 20)
g.DrawLine(Pens.AntiqueWhite, w, 0, w, height);
for (int h = 0; h <= height; h = h + 20)
g.DrawLine(Pens.AntiqueWhite, 0, h, width, h);
foreach (Vector v in datapoints)
{
g.FillEllipse(Brushes.Red, (int)v[0]  2, (int)v[1]  2, 4, 4);
g.DrawEllipse(Pens.Red, (int)v[0]  2, (int)v[1]  2, 4, 4);
}
foreach (Vector v in graph.Vertizes)
g.DrawEllipse(Pens.Indigo, (int)v[0]  2, (int)v[1]  2, 4, 4);
foreach (VoronoiEdge edge in graph.Edges)
{
try
{
if (!edge.IsPartlyInfinite)
{
g.DrawLine(Pens.Gray, (int)edge.VVertexA[0], (int)edge.VVertexA[1],
(int)edge.VVertexB[0], (int)edge.VVertexB[1]);
}
else
{
if (!edge.IsInfinite)
{
Vector VVertexB = edge.FixedPoint + edge.DirectionVector*edge.VVertexA.SquaredLength;
g.DrawLine(Pens.Blue, (int) edge.VVertexA[0], (int) edge.VVertexA[1],
(int) VVertexB[0], (int) VVertexB[1]);
}
else
{
Vector VVertexA = edge.FixedPoint 
edge.DirectionVector*Math.Sqrt(Math.Pow(width, 2) + Math.Pow(height, 2));
Vector VVertexB = edge.FixedPoint +
edge.DirectionVector*Math.Sqrt(Math.Pow(width, 2) + Math.Pow(height, 2));
g.DrawLine(Pens.Blue, (int)VVertexA[0], (int)VVertexA[1],
(int)VVertexB[0], (int)VVertexB[1]);
}
}
}
catch (Exception e)
{
}
}
return bmp;
}
public static Bitmap GetDelaunayTriangulation(int width, int height, HashSet<Vector> datapoints)
{
Bitmap bmp = new Bitmap(width, height);
VoronoiGraph graph = Fortune.ComputeVoronoiGraph(datapoints);
Graphics g = Graphics.FromImage(bmp);
foreach (Vector v in datapoints)
{
g.FillEllipse(Brushes.Red, (int)v[0]  2, (int)v[1]  2, 4, 4);
g.DrawEllipse(Pens.Red, (int)v[0]  2, (int)v[1]  2, 4, 4);
foreach (VoronoiEdge edge in graph.Edges)
if (edge.LeftData == v)
g.DrawLine(Pens.Black, (int)edge.LeftData[0], (int)edge.LeftData[1], (int)edge.RightData[0], (int)edge.RightData[1]);
}
return bmp;
}
}
}





Just a note to say I get the same thing unfortunately.





In most cases you have 2 infinite edges per area. Just find the crossing point of them and choose direction from this point to the VVertixA (or the midpoint between Left and Right points) of the edge.







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